Animated Solution for Physics - Gravitation: The planet Mars has two Moons, if one of them has a period 7 h, 30 min and an orbital radius of $9.0\times 10^3$ km. Find the mass of Mars. $\left[\text{Take, }\frac{4{\pi }^2}{G}=6\times 10^{11}{\text{ N}}^{-1}{\text{m}}^{-2}{\text{kg}}^2\right]$

Imagine you are an astronomer tasked with finding the mass of Mars. You can't exactly put a planet on a giant weighing scale! Instead, we use the delicate dance of celestial mechanics. By observing the orbit of one of its moons, we can deduce the mass of the central planet.

Let's start by setting up our physical parameters. We are given the orbital radius of the moon as $9.0\times 10^3\text{ km}$. In physics, we must always work in standard SI units to avoid catastrophic errors. So, we convert this to meters:

Next, we look at the time period of the orbit, which is $7\text{ h }30\text{ min}$. Converting this entirely into seconds gives us:

Balancing Forces

For the moon to maintain a stable, circular orbit around Mars, there must be a force constantly pulling it towards the center, preventing it from flying off into deep space. This is the centripetal force, and it is provided entirely by the gravitational force of attraction between Mars and the moon.

We can write this fundamental balance as:

Here, $M$ is the mass of Mars, $m$ is the mass of the moon, and $\omega$ is the angular velocity. Notice something beautiful? The mass of the moon ($m$) cancels out on both sides! This means the orbit of a satellite depends only on the mass of the central body, not on the satellite itself.

We know that angular velocity is related to the time period by the equation $\omega =\frac{2\pi }{T}$. Substituting this into our force balance equation, we get:

Our goal is to find the mass of Mars, $M$. Let's rearrange the equation to isolate $M$:

The problem generously provides the value for the entire constant term $\frac{4{\pi }^2}{G}=6\times 10^{11}{\text{ N}}^{-1}{\text{m}}^{-2}{\text{kg}}^2$. This saves us from dealing with the messy gravitational constant $G$ directly.

Now, we substitute our raw values into the rearranged equation. Don't rush the arithmetic here; let's write it out clearly:

Let's expand the powers. The cube of $9$ is $729$, and $(10^6{)}^3$ is $10^{18}$. For the denominator, $27000$ is $27\times 10^3$. Squaring this gives $729\times 10^6$.

The number $729$ cancels out perfectly from the numerator and denominator! This is a classic hallmark of a well-designed JEE problem. We are left with:

And there we have it! By simply observing how long it takes a moon to complete one orbit and measuring its distance, we have successfully weighed an entire planet.